For what value of $a$ is the sum of the squares of the numbers, constituting the solution of the system of equations $\;$ $3x - y = 2 - a$ $\;$ and $\;$ $x + 2y = a + 1$ $\;$ the least?
Given system of equations:
$3x - y = 2 - a$ $\;\;\; \cdots \; (1)$, and $\;$ $\;$ $x + 2y = a + 1$ $\;\;\; \cdots \; (2)$
Multiplying equation $(1)$ with $2$ gives
$6x - 2y = 4 - 2a$ $\;\;\; \cdots \; (3)$
Adding equations $(2)$ and $(3)$ gives
$7x = 5 - a$ $\implies$ $x = \dfrac{5 - a}{7}$
Substituting the value of $x$ in equation $(2)$ gives
$\dfrac{5 - a}{7} + 2y = a + 1$
i.e. $\;$ $2y = a + 1 - \dfrac{5 - a}{7}$
i.e. $\;$ $2y = \dfrac{7a + 7 - 5 + a}{7}$
i.e. $\;$ $2y = \dfrac{8a + 2}{7}$ $\implies$ $y = \dfrac{4a + 1}{7}$
Let $\;$ $s$ $\;$ be the sum of squares of the numbers constituting the solution of the given system of equations. Then,
$s = \left(\dfrac{5 - a}{7}\right)^2 + \left(\dfrac{4a + 1}{7}\right)^2$
i.e. $\;$ $s = \dfrac{25 + a^2 - 10 a + 16 a^2 + 1 + 8a}{49}$
i.e. $\;$ $s = \dfrac{17 a^2 - 2a + 26}{49}$ $\;\;\; \cdots \; (4)$
For $\;$ $s$ $\;$ to be minimum, $\;\;$ $\dfrac{ds}{da} = 0$
$\therefore \;$ We have from equation $(4)$,
$\dfrac{ds}{da} = \dfrac{17}{49} \times 2a - \dfrac{2}{49} + 0 = 0$
i.e. $\;$ $\dfrac{17 \times 2a}{49} = \dfrac{2}{49}$
i.e. $\;$ $a = \dfrac{1}{17}$